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PROFESSOR: Let's begin.
Today we're going to continue the discussion on Ito calculus.
I briefly introduced you to Ito's lemma last time,
but let's begin by reviewing it and stating it
in a slightly more general form.
Last time what we did was we did the quadratic variation
of Brownian motion, Brownian process.
We defined the Brownian process, Brownian motion,
and then showed that it has quadratic variation, which
can be written in this form-- dB square is equal to dt.
And then we used that to show the simple form of Ito's lemma,
which says that if f is a function on the Brownian
motion, then d of f is equal to f prime of dB_t
plus f double prime of dt.
This additional term was a characteristic of Ito calculus.
In classical calculus we only have this term,
but we have this additional term.
And if you remember, this happened exactly
because of this quadratic variation.
Let's review it, and let's do it in a slightly more general
form.
As you know, we have a function f
depending on two variables, t and x.
Now we're interested in-- we want to evaluate
our information on the function f(t, B_t).
The second coordinate, we're planning
to put in the Brownian motion there.
Then again, let's do the same analysis.
Can we describe d of f in terms of these differentiations?
To do that, deflect this, let me start from Taylor expansion.
f at a point t plus delta t, x plus delta x by Taylor
expansion for two variables is f of t of x plus partial of f
over partial of t at t comma x of delta t plus...
x.
That's the first-order terms.
Then we have the second-order terms.
Then the third-order terms, and so on.
That's just Taylor expansion.
If you look at it, we have a function f.
We want to look at the difference of f when we change
the first variable a little bit and the second variable
a little bit.
We start from f of t of x.
In the first-order terms, you take the partial derivative,
so take del f over del t, and then multiply
by the t difference.
Second term, you take the partial derivative
with respect to the second variable-- partial f
over partial x-- and then multiply by del x.
That much is enough for classical calculus.
But then, as we have seen before,
we ought to look at the second-order term.
So let's first write down what it is.
That's exactly what happened in Taylor expansion,
if you remember.
If you don't remember, just believe me.
This 1 over 2 times, take the second derivatives, partial.
Let's write it in terms of-- yes?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Oh, yeah, you're right.
Thank you.
Is it good now?
Let's write it as dt, all these deltas.
I'll just write like that.
I'll just not write down t and x.
And what we have is f plus del f over del t dt plus del
f over del x dx plus the second-order terms.
The only important terms-- first of all, these terms
are important.
But then, if you want to use x equals B of
t-- so if you're now interested in f t comma B of t.
Or more generally, if you're interested in f t plus dt,
f B_t plus d of B_t, then these terms are important.
If you subtract f of t of B_t, what you get
is these two terms.
Del f over del t dt plus del f over del
x-- I'm just writing this as a second variable
differentiation-- at dB_t.
And then the second-order terms.
Instead of writing it all down, dt square is insignificant,
and dt comma-- dt times dB_t also is insignificant.
But the only thing that matters will be this one.
This is dB_t square, which you saw is equal to dt.
From the second-order term, we'll have this term surviving.
1 over 2 partial f over partial x second derivative, of dt.
That's it.
If you rearrange it, what we get is
partial f over partial t plus 1/2 this plus--
and that's the additional term.
If you ask me why these terms are not important
and this term is important, I can't really say it rigorously.
But if you think about dB_t square equals dt, then d times
B_t is kind of like square root of dt.
It's not a good notation, but if you
do that-- these two terms are significantly smaller than dt
because you're taking a power of it.
dt square becomes a lot smaller than dt, dt to the 3/2
is a lot smaller than dt.
But this one survives because it's equal to dt here.
That's just the high-level description.
That's a slightly more sophisticated form
of Ito's lemma.
Let me write it down here.
And let's just fix it now.
If f of t of B_t-- that's d of f is equal to-- Any questions?
Just remember, from the classical calculus term,
we're only adding this one term there.
Yes?
AUDIENCE: Why do we have x there?
PROFESSOR: Because the second variable is supposed to be x.
I don't want to write down partial derivative with respect
to a Brownian motion here because it doesn't look good.
It just means, take the partial derivative with respect
to the second term.
So just view this as a function f of t of x,
evaluate it, and then plug in x equal B_t in the end,
because I don't want to write down partial B_t here.
Other questions?
Consider a stochastic process X of t such that d of X
is equal to mu times d of t plus sigma times d of B_t.
This is almost like a Brownian motion,
but you have this additional term.
This is called a drift term.
Basically, this happens if X_t is equal to mu*t plus sigma
of B_t.
Mu and sigma are constants.
From now on, what we're going to study
is stochastic process of this type, whose difference
can be written in terms of drift term and the Brownian motion
term.
We want to do a slightly more general form
of Ito's lemma, where we want f of t of X_t here.
That will be the main object of study.
I'll finally state the strongest Ito's lemma
that we're going to use.
f is some smooth function and X_t is a stochastic process
like that.
X_t satisfies...
where B_t is a Brownian motion.
Then df of t, X_t can be expressed
as-- it's just getting more and more complicated.
But it's based on this one simple principle, really.
It all happened because of quadratic variation.
Now I'll show you why this form deviates from this form when
we replace B to an X.
Remember here all other terms didn't matter,
that the only term that mattered was partial square of f...
of dx square.
To prove this, note that df is partial f over partial t
dt plus partial f over partial x d of X_t
plus 1/2 of d of x squared.
Just exactly the same, but I've replaced the dB-- previously,
what we had dB, I'm replacing to dX.
Now what changes is dX_t can be written like that.
If you just plug it in, what you get here
is partial f over partial x mu dt plus sigma of dB_t.
Then what you get here is 1/2 of partials
and then mu plus sigma dB_t square.
Out of those three terms here we get mu square dt square
plus 2 times mu sigma d mu dB plus sigma square dB square.
Only this was survives, just as before.
These ones disappear.
And then you just collect the terms.
So dt-- there's one dt here.
There's mu times that here, and that one will become a dt.
It's 1/2 of sigma square partial square...
of dt.
And there's only one dB_t term here.
Sigma-- I made a mistake, sigma.
This will be a form that you'll use the most,
because you want to evaluate some stochastic process--
some function that depends on time
and that stochastic process.
You want to understand the difference, df.
The X would have been written in terms
of a Brownian motion and a drift term,
and then that's the Ito lemma for you.
But if you want to just-- if you just
see this for the first time, it just looks too complicated.
You don't understand where all the terms are coming from.
But in reality, what it's really doing
is just take this Taylor expansion.
Remember these two classical terms,
and remember that there's one more term here.
You can derive it if you want to.
Really try to know where it all comes from.
It all started from this one fact, quadratic variation,
because that made some of the second derivative survive,
and because of those, you get these kind
of complicated terms.
Questions?
Let's do some examples.
That's too much.
Sorry, I'm going to use it a lot, so let me record it.
Example number one.
Let f of x be equal to x square, and then you
want to compute d of f at B_t.
I'll give you three minutes just to try a practice.
Did you manage to do this?
It's a very simple example.
Assume it's just the function of two variables,
but it doesn't depend on t.
You don't have to do that, but let me just do that.
Partial f over partial t is 0.
Partial f over partial x is equal to 2x,
and the second derivative equal to 2 at t, x.
Now we just plug in t comma B_t, and what
you have is mu equals 0, sigma equals 1,
if you want to write it in this formula.
What you're going to have is 2 times B_t of dB_t
plus 1 over 2 times 2dt.
If you write it down.
You can either use these parameters
and just plug in each of them to figure it out.
Or a different way to do it is really
write down, remember the proof.
This is partial f over partial t dt
plus partial f over partial x dx plus 1/2-- remember this one.
And x is dB_t here.
That one is 0, that one was 2x, so 2B_t dB_t.
Use it one more time, so you get dt.
Make sense?
Let's do a few more examples.
And you want to compute d of f at t comma B of t.
Let's do it this time.
Again, partial f over partial t dt plus partial f
over partial x dB_t.
That's the first-order terms.
The second-order term is 1/2 partial square f
over partial x square of dB_t square, which is equal to dt.
Let's do it.
Partial f over partial t, you get mu times f.
This one is just equal to mu times f.
Maybe I'm going too quick.
Mu times e to the mu t plus dx, dt.
Partial f over partial x is sigma times e
to the mu t plus dx, and then dB_t
plus-- if you take the second derivative,
you do that again, what you get is
1/2, and then sigma square times e to the mu t plus dx, dt.
Yes?
AUDIENCE: In the original equation that you just wrote,
isn't it 1/2 times sigma squared,
and then the second derivative?
Up there.
PROFESSOR: Here?
AUDIENCE: Yes.
PROFESSOR: 1/2?
AUDIENCE: Times sigma squared.
PROFESSOR: Oh, sigma-- OK, that's a good question.
But that sigma is different.
That's if you plug in X_t here.
If you plug in X_t where X_t is equal to mu
prime dt plus sigma prime d of B_t,
then that sigma prime will become a sigma prime square
here.
But here the function is mu and sigma,
so maybe it's not a good notation.
Let me use a and b here instead.
The sigma here is different from here.
AUDIENCE: Yeah, that makes a lot more sense.
PROFESSOR: If you replace a and b,
but I already wrote down all mu's and sigma's.
That's a good point, actually.
But that's when you want to consider
a general stochastic process here
other than Brownian motion.
But here it's just a Brownian motion,
so it's the most simple form.
And that's what you get.
Mu plus 1/2 sigma square-- and these are just all f itself.
That's the good thing about exponential.
f times dt plus sigma times d of B_t.
Make sense?
And there's a reason I was covering this example.
It's because-- let's come back to this question.
You want to model stock price using Brownian motion, Brownian
process, S of t.
But you don't want S_t to be a Brownian motion.
What you want is a percentile difference
to be a Brownian motion, so you want this percentile difference
to behave like a Brownian motion with some variance.
The question was, is S_t equal to e to the sigma times B of t
in this case?
And I already told you last time that no, it's not true.
We can now see why it's not true.
Take this function, S_t equals e to the sigma
B_t, that's exactly where mu is equal to 0 here.
What we got here was d of S_t, in this case, is equal to-- mu
is 0, so we get 1/2 of sigma square times dt plus sigma
times d of B_t.
We originally were targeting sigma times dB_t,
but we got this additional term which we
didn't want in the first hand.
In other words, we have this drift.
I wasn't really clear in the beginning,
but our goal was to model stock price
where the expected value is 0 at all times.
Our guess was to take e to the sigma B_t,
but it turns out that in this case
we have a drift, if you just take
natural e to the sigma of B_t.
To remove that drift, what you can do
is subtract that term somehow.
If you can get rid of that term then you can see,
if you add this mu to be minus 1 over 2 sigma square,
you can remove that term.
That's why it doesn't work.
So instead use S of t equals e to the minus 1 over 2 sigma
square t plus sigma of B_t.
That's the geometric Brownian motion without drift.
And the reason it has no drift is because of that.
If you actually do the computation,
the dt term disappears.
Question?
So far we have been discussing differentiation.
Now let's talk about integration.
Yes?
AUDIENCE: Could you we do get this solution as [INAUDIBLE].
Could you also describe what it means?
What does it mean, this solution of B_t?
Does that mean if we have a sample path B_t,
then we could get a sample path for S_t?
PROFESSOR: Oh, what this means, yes.
Whenever you have the B_t value, just at each time
take the exponential value.
Because-- why we want to express this in terms of a Brownian
motion is, for Brownian motion we
have a pretty good understanding.
It's a really good process you understand fairly well,
and you have good control on it.
But the problem is you want to have a process whose percentile
difference behaves like a Brownian motion.
And this gives you a way of describing it
in terms of Brownian motion, as an exponential function of it.
Does that answer your question?
AUDIENCE: Right, distribution means
that if we have a sample path B_t,
that would be the corresponding sample path for S of t?
Is it a pointwise evaluation?
PROFESSOR: That's a good question, actually.
Think of it as a pointwise evaluation.
That is not always correct, but for most
of the things that we will cover,
it's safe to think about it that way.
But if you think about it path-wise all the time,
eventually it fails.
But that's a very advanced topic.
So what this question is, basically
B_t is a probability space, it's a probability distribution
over paths.
For this equation, if you just look at it, it looks right,
but it doesn't really make sense,
because B_t-- if it's a probability distribution, what
is e to the B_t?
Basically, what it's saying is B_t
is a probability distribution over paths.
If you take omega according to-- a path according
to the Brownian motion sample probability distribution,
and for this path it's well defined, this function.
So the probability density function of this path
is equal to the probability density function of e
to the whatever that is in this distribution.
Maybe it confused you more.
Just consider this as some path, some well-defined function,
and you have a well-defined function.
Integral definition.
I will first give you a very, very stupid definition
of integration.
We say that we define F as the integration...
if d of F is equal to f dB_t plus-- we
define it as an inverse of differentiation.
Because differentiation is now well-defined--
we just defined integration as the inverse of it,
just as in classical calculus.
So far, it doesn't have that good meaning,
other than being an inverse of it,
but at least it's well-defined.
The question is, does it exist?
Given f and g, does it exist, does integration always exist,
and so on.
There's lots of questions to ask,
but at least this is some definition.
And the natural question is, does there exist a Riemannian
sum type description?
That means-- if you remember how we defined integral
in calculus, you have a function f, integration
of f from a to b according to the Riemannian sum description
was, you just chop the interval into very fine pieces,
a_0, a_1, a_2, a_3, dot, dot, dot--
and then sum the area of these boxes, and take the limit.
And this is the limit of Riemannian sums.
Slightly more, if you want, it's the limit as n goes to infinity
of the function 1 over n times the sum of i equal zero to t--
I'll just call it 0 to b-- f of t*b over n minus f of t minus 1
over n.
Does this ring a bell?
Question?
AUDIENCE: [INAUDIBLE]
PROFESSOR: No, you're right.
Good point, no we don't.
Thanks.
Does integral defined in this way
have this Riemannian sum type description, is the question.
So keep that in mind.
I will come back to this point later.
In fact, it turns out to be a very deep question
and very important question, this question,
because if you remember like I hope you remember,
in the Riemannian sum, it didn't matter which point you
took in this interval.
That was the whole point.
You have the function.
In the interval a_i to a_(i+1), you take any point
in the middle and make a rectangle according to that
point.
And then, no matter which point you take,
when you go to the limit, you had exactly the same sum
all the time.
That's how you define the limit.
But what's really interesting here
is that it's no longer true.
If you take the left point all the time,
and you take the right point all the time,
the two limits are different.
And again, that's due to the quadratic variation,
because that much of variance can accumulate over time.
That's the reason we didn't start with Riemannian sum type
definition of integral.
But I'll just make one remark.
Ito integral is the limit of Riemannian sums
when always take the leftmost point of each interval.
So you chop down this curve at-- the time interval
into pieces, and for each rectangle,
pick the leftmost point, and use it as a rectangle.
And you take the limit.
That will be your Ito integral defined.
It will be exactly equal to this thing, the inverse of our Ito
differentiation.
I won't be able to go into detail.
What's more interesting is instead,
what happens if you take the rightmost point all the time,
you get an equivalent theory of calculus.
It's just like Ito's calculus.
It looks really, really similar and it's coherent itself,
so there is no logical flaw in it.
It all makes sense, but the only difference
is instead of a plus in the second-order term,
you get minuses.
Let me just make this remark, because it's just
a theoretical part, this thing, but I think it's really cool.
Remark-- there's this and equivalent version.
Maybe equivalent is not the right word,
but a very similar version of Ito
calculus such that basically, what
it says is d B_t square is equal to minus dt.
Then that changed a lot of things.
But this part, it's not that important.
Just cool stuff.
Let's think about this a little bit more, this fact.
Taking the leftmost point all the time
means if you want to make a decision for your time
interval-- so at time t of i and time t of i plus 1,
let's say it's the stock price.
You want to say that you had so many stocks in this time
interval.
Let's say you had so many stocks in this time interval
according to the values between this and this.
In real world, your only choice you have is you
have to make the decision at time t of i.
Your choice cannot depend on the future time.
You can't suddenly say, OK, in this interval the stock
price increased a lot, so I'll assume
that I had a lot of stocks in this interval.
In this interval, I knew it was going to drop,
so I'll just take the rightmost interval.
I'll assume that I only had this many stock.
You can't do that.
Your decision has to be based on the leftmost point,
because the time.
You can't see the future.
And the reason Ito's calculus works well in our setting is
because of this fact, because it has inside it the fact that you
cannot see the future.
Every decision is made based on the leftmost time.
If you want to make a decision for your time interval,
you have to do it in the beginning.
That intuition is hidden inside of the theory,
and that's why it works so well.
Let me reiterate this part a little bit more.
It's the definition of these things
where you're only allowed to-- at time t,
you're only allowed to use the information up to time t.
Definition: delta t is an adapted process-- sorry,
adapted to another stochastic process X_t--
if for all values of time variables
delta t depends only on X_0 up to X_t.
There's a lot of vague statements inside here,
but what I'm trying to say is just
assume X is the Brownian motion underlying stock price.
Your stock is changing.
You want to come up with a strategy,
and you want to say that mathematically
this strategy makes sense.
And what it's saying is if your strategy
makes your decision at time t is only
based on the past values of your stock price,
then that's an adapted process.
This defines the processes that are reasonable,
that cannot see future.
And these are all-- in terms of strategy,
if delta_t is a portfolio strategy,
these are the only meaningful strategies that you can use.
And because of what I said before, because we're always
taking the leftmost point, adaptive
processes just also fit very well with Ito's calculus.
They'll come into play altogether.
Just a few examples.
First, a very stupid example.
X_t is adapted to X_t.
Of course, because at time, X_t really
depends on only X_t, nothing else.
Two, X_(t+1) is not adapted to X_t.
This is maybe a little bit vague,
so we'll call it Y_t equals X_(t+1).
Y_t is the value at t plus 1, and it's not based
on the values up to time t.
Just a very artificial example.
Another example, delta t equals minimum...
is adapted.
And I'll let you think about it.
The fourth is quite interesting.
Suppose T is fixed, some large integer,
or some large real number.
Then you let delta t to be the maximum where X of s, where...
It's not adapted.
What is this?
This means at time T, I'm going to take at it
this value, the maximum of all value
inside this part, the future.
This refers to the future.
It's not an adapted process.
Any questions?
Now we're ready to talk about the properties
of Ito's integral.
Let's quickly review what we have.
First, I defined Ito's lemma-- that means differentiation
in Ito calculus.
Then I defined integration using differentiation-- integration
was an inverse operation of the differentiation.
But this integration also had an alternative description
in terms of Riemannian sums, where
you're taking just the leftmost point as the reference
point for each interval.
And then, as you see, this naturally
had this concept of using the leftmost point.
And to abstract that concept, we've
come up with this adapted process, very natural process,
which is like the real-life procedures,
real-life strategies we can think of.
Now let's see what happens when you
take the integral of adapted processes.
Ito integral has really cool properties.
The first thing is about normal distribution.
B_t has normal distribution of 0 up to t.
So your Brownian motion at time t
has normal distribution with 0, t.
That means if your stochastic process is some constant times
B of t, of course, then you have 0 and c square t.
It's still a normal variable.
That means if you integrate, that's
the integration of some sigma.
That's the integration of sigma of dB_t.
If sigma is a fixed constant, when
you take the Ito integral of sigma times
dB_t, this constant, at each time
you get a normal distribution.
And this is like saying the sum of normal distribution
is also normal distribution.
It has this hidden fact, because integral
is like sum in the limit.
And this can be generalized.
If delta t is a process depending only on the time
variable-- so it does not depend on the Brownian motion-- then
the process X of t equals the integration of delta t dB_t
has normal distribution at all time, just like this.
We don't know the exact variance yet;
the variance will depend on the sigmas.
But still, it's like a sum of normal variables,
so we'll have normal distribution.
In fact, it just gets better and better.
The second fact is called Ito isometry.
That was cool.
Can we compute the variance?
Yes?
AUDIENCE: Can you put that board up?
PROFESSOR: Sure.
AUDIENCE: Does it go up?
PROFESSOR: This one doesn't go up.
That's bad.
I wish it did go up.
This has a name called Ito isometry.
Can be used to compute the variance.
B_t is a Brownian motion, delta t
is adapted to a Brownian motion.
Then the expectation of your Ito integral--
that's the Ito integral of your adapted process.
That's the variance-- we take the square of it--
is equal to something cool.
The square just comes in.
Quite nice, isn't it?
I won't prove it, but let me tell you why.
We already saw this phenomenon before.
This is basically quadratic variation.
And the proof also uses it.
If you take delta s equals to 1-- sorry,
I was using Korean-- 1 at all time, then what we have is
here you get a Brownian motion, B_t.
So on the left you get like expectation of B_t square,
and on the right, what you get is t.
Because when delta s is equal to 1
at all time, when you have to get from 0 to t you get t,
and you have t on the right hand side.
That's what it's saying.
And that was the content of quadratic variation,
if you remember.
We're summing the squares-- maybe not exactly this,
but you're summing the squares over small intervals.
So that's a really good fact that you can
use to compute the variance.
You have an Ito integral, you know the square,
can be computed this simple way.
That's really cool.
And one more property.
This one will be really important.
You'll see it a lot in future lectures.
It's that when is Ito integral a martingale?
What's a martingale?
Martingale meant if you have a stochastic process,
at any time t, whatever happens after that, the expected value
at time t is equal to 0.
It doesn't have any natural tendency to go up or go down.
No matter which point you stop your process
and you see your future, it doesn't have a natural tendency
to go up or go down.
In formal language, it can be defined as where F_t
is the events X_0 up to X_t.
So if you take the conditional expectation
based on whatever happened up to time t,
that expectation will just be whatever value
you have at that time.
Intuitively, that just means you don't have any natural tendency
to go up or go down.
Question is, when is an Ito integral a martingale?
Adapted to B of t, then...
is a martingale.
As long as g is not some crazy function,
as long as g is reasonable-- one way can be reasonable if its
L^2-norm is bounded.
If you don't know what it means, you can safely ignore it.
Basically, if g doesn't-- it's not a crazy function if it
doesn't grow too fast, then in most cases this integral is
always a martingale.
If you flip it-- remember, integral
was defined as the inverse of differentiation.
So if dX_t is equal to some function mu, that
depends on both t and B_t, times dt, plus sigma
of dB_t, what this means is X_t is a martingale
if that is 0 at all time, always.
And if it's not 0, you have a drift,
so it's not a martingale.
That gives you some classification.
Now, if you look at a differential equation
of this stochastic-- this is called
a stochastic differential equation-- if you know
stochastic process, if you look at a stochastic differential
equation, if it doesn't have a drift term, it's a martingale.
If it has a drift term, it's not a martingale.
That'll be really useful later, so try to remember it.
The whole point is when you write down
a stochastic process in terms of something times dt,
something times dB_t, really this term
contributes towards the tendency, the slope of whatever
is going to happen in the future.
And this is like the variance term.
It adds some variance to your stochastic process.
But still, it doesn't add or subtract value over time,
it fairly adds variation.
Remember that.
That's very important fact.
You're going to use it a lot.
For example, you're going to use it for pricing theory.
In pricing theory, you come up with this stochastic process
or some strategy.
You look at its value.
Let's say X_t is your value of your portfolio over time.
If that portfolio has-- then you match it with your financial--
let me go over it slowly again.
First you have a financial derivative, like option
of a stock.
Then you have your portfolio strategy.
Assume that you have some strategy that,
at the expiration time, gives you
the exact value of the option.
Now you look at the difference between these two
stochastic processes.
Basically what the thing is, when your variance goes to 0,
your drift also has to go to 0.
So when you look at the difference,
if you can somehow get rid of this variance term, that
means no matter what you do, that
will govern the value of your portfolio.
If it's positive, that means you can always make money,
because there's no variance.
Without variance, you make money.
That's called arbitrage, and you cannot have that.
But I won't go into further detail
because Vasily will cover it next time.
But just remember that flavor.
So when you write something down in a stochastic differential
equation form, that term is a drift term,
that term is a variance term.
And if you don't have drift, it's a martingale.
That is very important.
Any questions?
That's kind of the basics of Ito calculus.
I will give you some exercises on it,
mostly just basic computation exercises, so that you'll
get familiar with it.
Try to practice it.
And let me cover one more thing called Girsanov theorem.
It's related, but these are really
basics of the Ito calculus, so if you
have any questions on this, please ask me
right now before I move on to the next topic.
The last thing I want to talk about today.
Here is an underlying question.
Suppose you have two Brownian motions.
This is without drift.
And you have another B tilde, Brownian motion with drift.
These are two probability distributions over paths.
According to B_t, you're more likely to have
some Brownian motion that has no drift.
That's a sample path.
According to B tilde, you have some drift.
Your Brownian motion will--
A typical path will follow this line and will follow that line.
The question is this-- can we switch
from this distribution to this distribution
by a change of measure?
Can we switch between the two measures
to probability distributions by a change of measure?
Let me go a little bit more what it really means.
Assume that you're just looking at a Brownian motion
from time 0 up to time t, some fixed time interval.
Then according to B_t, let's say this is a sample path omega.
You have some probability of omega-- this is a p.d.f.
given by this Brownian motion B. And then you
have another p.d.f., P tilde of omega, which is a p.d.f.
given by B of t.
The question is, does there exist a Z
depending on omega such that P of omega
is equal to Z times P tilde?
Do you understand the question?
Clearly, if you just look at it, they're quite different.
The path that you get according to the distributions
are quite different.
It's not clear why we should expect it at all.
You'll see the answer soon.
But let me discuss all this in a different context.
Just forget about all the Brownian motion and everything
just for a moment.
In this concept, changing from one probability distribution
to another distribution, it's a very important concept
in analysis and probability just in general, theoretically.
And there's a name for this Z, for this changing measure.
If Z exists, it's called the Radon-Nikodym derivative.
Before doing that, let me talk a little bit more.
Suppose P is a probability distribution over omega.
It's a probability distribution.
So this is some set, and P describes the probability
that you have each element in the set.
And you have another probability distribution, P tilde.
We define P and P tilde to be equivalent if the probability
that A is greater than zero if and only if...
For all...
These probability distributions describe the probability
of the subsets.
Think about a very simple case.
Sigma is equal to 1, 2, and 3.
P gives 1/3 probability to 1, 1/3 probability
to 2, 1/3 probability to 3.
P tilde gives 2/3 probability to 3, 1 over 6 probability
to 2, 1 over 6 probability to 3.
We have two probability distribution over some space.
They are equivalent if, whenever you
take a subset of your background set-- let's say 1, 2.
When A is equal to 1, 2, according
to probability distribution P, the probability
you fall into this set A is equal to 2/3.
According to P tilde, you have 5/6.
They're not the same.
The probability itself is not the same,
but this condition is satisfied when it's 0.
And when it's not 0, it's not 0.
And you can just check that it's always true, because they're
all positive probabilities.
On the other hand, if you take instead, say,
1/3 and 0, now you take your A to be 3.
Then you have 1/3 equal to 0.
This means, according to probability distribution P,
there is some probability that you'll get 3.
But according to probability distribution P tilde,
you don't have any probability of getting 3.
So they're not equivalent in this case.
If you think about it, then it's really clear.
The theorem says-- this is a very important theorem
in analysis, actually.
The theorem-- there exists a Z such that P of omega is equal
to...
If and only if P and P tilde are equivalent.
You can change from one probability measure
to another probability measure just
in terms of multiplication, if and only if they're equivalent.
And you can see that it's not the case for this
when they're not equivalent.
You can't make a zero probability to 1/3 probability
by multiplication.
So in the finite world this is very just intuitive theorem,
but what this is saying is it's true for all probability
spaces.
And these are called the Radon-Nikodym derivative.
Our question is, are these two Brownian motions equivalent?
The paths that this Brownian motion without drift
takes and the Brownian motion with drift
takes, are they kind of the same but just skewed
in distribution, or are they really fundamentally different?
That's the question.
And what Girsanov's theorem says is that they are equivalent.
To me, it came as a little bit non-intuitive.
I would imagine that it's not equivalent, these two.
These paths have a very natural tendency.
As it goes to infinity, these paths and these paths
will really look a lot different,
because when you go really, really far,
the paths which have drift will be just really
close to your line mu of t, while the paths which
don't have drift will be really close to the x-axis.
But still, they are equivalent.
You can change from one to another.
I'll just state that theorem without proof.
And this will also be used in pricing theory.
I'm not an expert enough to tell why,
but basically what it's saying is,
you switch some stochastic process
into a stochastic process without drift,
thus making it into a martingale.
And martingale has a lot of meaning in pricing theory,
as you'll see.
This also has application.
That's why I'm trying to cover it, although it's
quite a technical theorem.
Try to remember at least a statement and the spirit
of what it means.
It just means these two are equivalent,
you can change from one to another
by a multiplicative function.
Let me just state it in a simple form.
GUEST SPEAKER: If I could just interject a comment.
PROFESSOR: Sure.
GUEST SPEAKER: With these changes of measure,
it turns out that all of these theories with continuous time
processes should have an interpretation if you've
discretized time, and should consider
sort of a finer and finer discretization of the process.
And with this change of measure, if you consider problems
in discrete stochastic processes like random walks,
basically how-- say if you're gambling against a casino
or against another player, and you
look at how your winnings evolve as a random walk,
depending on your odds, your odds
could be that you will tend to lose.
So there's basically a drift in your wealth
as this random process evolves.
You can transform that process, basically by taking out
your expected losses, to a process which
has zero change in expectation.
And so you can convert these gambling problems
where there's drift to a version where the process, essentially,
has no drift and is a martingale.
And the martingale theory in stochastic process courses
is very, very powerful.
There's martingale convergence theorems.
So you know that the limit of the martingale
is-- there's a convergence of the process,
and that applies here as well.
PROFESSOR: You will see some surprising applications.
GUEST SPEAKER: Yeah.
PROFESSOR: And try to at least digest the statement.
When the guest speaker comes and says by Girsanov theorem,
they actually know what it is.
There's a spirit.
This is a very simple version.
There's a lot of complicated versions,
but let me just do it.
So P is a probability distribution over paths
from [0, T] to the infinity.
What this means is just paths from that-- stochastic process
defined from time 0 to time T. These
are paths defined by a Brownian motion with drift mu.
And then P tilde is a probability distribution
defined by Brownian motion without drift.
Then P and P tilde are equivalent.
Not only are they equivalent, we can actually
compute their Radon-Nikodym derivative.
And the Radon-Nikodym derivative Z
which is defined as T of-- which we denote like this
has this nice form.
That's a nice closed form.
Let me just tell you a few implications of this.
Now, assume you have some, let's say, value
of your portfolio over time.
That's the stochastic process.
And you measure it according to this probability distribution.
Let's say it depends on some stock price
as the stock price is modeled using a Brownian
motion with drift.
What this is saying is, now, instead
of computing this expectation in your probability space--
so this is defined over the probability space P,
our sigma-- (omega, P) defined by this probability
distribution.
You can instead compute it in-- you
can compute as expectation in a different probability space.
You transform the problems about Brownian motion with drift
into a problem about Brownian motion without a drift.
And the reason I have Z tilde instead of Z
here is because I flipped.
What you really should have is Z tilde here as expectation of Z.
If you want to use this Z.
I don't expect you to really be able to do computations
and do that just by looking at this theorem once.
Just really trying to digest what it means
and understand the flavor of it, that you can transform
problems in one probability space
to another probability space.
And you can actually do that when the two distributions are
defined by Brownian motions when one has drift and one
doesn't have a drift.
How we're going to use it is we're
going to transform a non-martingale process
into a martingale process.
When you change into martingale it
has very good physical meanings to it.
That's it for today.
And you only have one more math lecture remaining
and maybe one or two homeworks but if you have two,
the second one won't be that long.
And you'll have a lot of guest lectures, exciting guest
lectures, so try not to miss them.